Parsing & Context-Free Grammars CSE 413 Autumn 2008 Parsing & Context-Free Grammars 6/6/2018

Agenda for Today Parsing overview Context free grammars Ambiguous grammars 6/6/2018

Parsing The syntax of most programming languages can be specified by a context-free grammar (CGF) Parsing: Given a grammar G and a sentence w in L(G ), traverse the derivation (parse tree) for w in some standard order and do something useful at each node The tree might not be produced explicitly, but the control flow of a parser corresponds to a traversal 6/6/2018

Old Example a = 1 ; if ( a + 1 ) b = 2 ; G w program program statement program ::= statement | program statement statement ::= assignStmt | ifStmt assignStmt ::= id = expr ; ifStmt ::= if ( expr ) statement expr ::= id | int | expr + expr id ::= a | b | c | i | j | k | n | x | y | z int ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Old Example G program a = 1 ; if ( a + 1 ) b = 2 ; program statement statement ifStmt statement assignStmt assignStmt expr id expr expr expr id expr int id int int w 6/6/2018

“Standard Order” For practical reasons we want the parser to be deterministic (no backtracking), and we want to examine the source program from left to right. (i.e., parse the program in linear time in the order it appears in the source file) 6/6/2018

Common Orderings Top-down Bottom-up Start with the root Traverse the parse tree depth-first, left-to-right (leftmost derivation) LL(k) Bottom-up Start at leaves and build up to the root Effectively a rightmost derivation in reverse(!) LR(k) and subsets (LALR(k), SLR(k), etc.) 6/6/2018

“Something Useful” At each point (node) in the traversal, perform some semantic action Construct nodes of full parse tree (rare) Construct abstract syntax tree (common) Construct linear, lower-level representation (more common in later parts of a modern compiler) Generate target code or interpret on the fly (1-pass compiler & interpreters; not common in production compilers – but what we will do for our project) 6/6/2018

Aside: ASTs Essential Structure Only program ::= statement | program statement statement ::= assignStmt | ifStmt assignStmt ::= id = expr ; ifStmt ::= if ( expr ) statement expr ::= id | int | expr + expr id ::= a | b | c | i | j | k | n | x | y | z int ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Aside: ASTs Essential Structure Only G program a = 1 ; if ( a + 1 ) b = 2 ; w 6/6/2018

Context-Free Grammars Formally, a grammar G is a tuple <N,Σ,P,S> where: N a finite set of non-terminal symbols Σ a finite set of terminal symbols P a finite set of productions A subset of N × (N  Σ )* S the start symbol, a distinguished element of N If not specified otherwise, this is usually assumed to be the non-terminal on the left of the first production 6/6/2018

Standard Notations a, b, c elements of Σ w, x, y, z elements of Σ* A, B, C elements of N X, Y, Z elements of N Σ , ,  elements of (N Σ )* A  or A ::=  if <A, > in P 6/6/2018

Derivation Relations (1)  A  =>    iff A ::=  in P derives A =>* w if there is a chain of productions starting with A that generates w transitive closure 6/6/2018

Derivation Relations (2) w A  =>lm w   iff A ::=  in P derives leftmost  A w =>rm   w iff A ::=  in P derives rightmost Parsers normally work with only leftmost or rightmost derivations – not random orderings 6/6/2018

Languages For A in N, L(A) = { w | A =>* w } i.e., set of strings (words, terminal symbols) generated by nonterminal A If S is the start symbol of grammar G, define L(G ) = L(S ) 6/6/2018

Reduced Grammars Grammar G is reduced iff for every production A ::=  in G there is some derivation S =>* x A z => x  z =>* xyz i.e., no production is useless Convention: we will use only reduced grammars 6/6/2018

Example Top down, Leftmost derivation for: a = 1 + b ; program ::= statement | program statement statement ::= assignStmt | ifStmt assignStmt ::= id = expr ; ifStmt ::= if ( expr ) stmt expr ::= id | int | expr + expr id ::= a | b | c | i | j | k | n | x | y | z int ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Example Top down, Leftmost derivation for: a = 1 + b ; 6/6/2018

Example Grammar S ::= aABe A ::= Abc | b B ::= d Top down, leftmost derivation of: abbcde 6/6/2018

Ambiguity Grammar G is unambiguous iff every w in L(G ) has a unique leftmost (or rightmost) derivation Fact: either unique leftmost or unique rightmost implies the other A grammar without this property is ambiguous Note that other grammars that generate the same language may be unambiguous We need unambiguous grammars for parsing 6/6/2018

Example: Ambiguous Grammar for Arithmetic Expressions expr ::= expr + expr | expr - expr | expr * expr | expr / expr | int int ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Exercise: show that this is ambiguous How? Show two different leftmost or rightmost derivations for the same string Equivalently: show two different parse trees for the same string 6/6/2018

Example (cont) Give a leftmost derivation of 2+3*4 and show the parse tree 6/6/2018

Example (cont) Give a different leftmost derivation of 2+3*4 and show the parse tree 6/6/2018

Another example Give two different derivations of 5+6+7 6/6/2018

What’s going on here? This grammar has no notion of precedence or associatively Standard solution Create a non-terminal for each level of precedence Isolate the corresponding part of the grammar Force the parser to recognize higher precedence subexpressions first ??? 6/6/2018

Classic Expression Grammar expr ::= expr + term | expr – term | term term ::= term * factor | term / factor | factor factor ::= int | ( expr ) int ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 6/6/2018

Check: Derive 2 + 3 * 4 6/6/2018

Check: Derive 5 + 6 + 7 Note interaction between left- vs right-recursive rules and resulting associativity 6/6/2018

Check: Derive 5 + (6 + 7) 6/6/2018

Another Classic Example Grammar for conditional statements stmt ::= if ( cond ) stmt | if ( cond ) stmt else stmt | assign Exercise: show that this is ambiguous How? 6/6/2018

One Derivation if ( cond ) if ( cond ) stmt else stmt stmt ::= if ( cond ) stmt | if ( cond ) stmt else stmt | assign One Derivation if ( cond ) if ( cond ) stmt else stmt 6/6/2018

Another Derivation if ( cond ) if ( cond ) stmt else stmt stmt ::= if ( cond ) stmt | if ( cond ) stmt else stmt | assign Another Derivation if ( cond ) if ( cond ) stmt else stmt 6/6/2018

Solving if Ambiguity Fix the grammar to separate if statements with else from if statements with no else Done in original Java reference grammar Adds lots of non-terminals Need productions for things like “while statement that has unmatched if” and “while statement with matched if”, etc. etc. etc. Use some ad-hoc rule in parser “else matches closest unpaired if” 6/6/2018

Parser Tools and Operators Most parser tools can cope with ambiguous grammars Makes life simpler if used with discipline Typically one can specify operator precedence & associativity Allows simpler, ambiguous grammar with fewer nonterminals as basis for generated parser, without creating problems 6/6/2018

Parser Tools and Ambiguous Grammars Possible rules for resolving other problems Earlier productions in the grammar preferred to later ones Longest match used if there is a choice Parser tools normally allow for this But be sure that what the tool does is really what you want 6/6/2018

Or… If the parser is hand-written, either fudge the grammar or the parser or cheat where it helps. 6/6/2018